Linear Combinations and description geometrically

kanderson

Member
Ok give me a break, this is my first lesson in my new linear algebra book. Seems fairly straightforward but a little befuddled as to whether I am doing this. The question states "Describe geometrically (line, plane, or all of R^3) all linear combinations of...."
Then I have a matrix v = [1 2 3] w = [3 6 9]...... I do this.

cv-dw [-2 -4 -6] I believe it is R^3
v+w [4 8 11] Also R^3

If you could offer if I am still befuddled, [1 0 0] and [0 2 3]

forgot my proof....cv + dw = c[1 1 0]+d[0 1 1] it turns out as cv+dw = [c c+d d]

Need to start learning latex -.-

(Maybe I can get ahead of Jameson and his linear algebra *snickers*) Last edited:

topsquark

Well-known member
MHB Math Helper
Ok give me a break, this is my first lesson in my new linear algebra book. Seems fairly straightforward but a little befuddled as to whether I am doing this. The question states "Describe geometrically (line, plane, or all of R^3) all linear combinations of...."
Then I have a matrix v = [1 2 3] w = [3 6 9]...... I do this.

cv-dw [-2 -4 -6] I believe it is R^3
I'm rather confused here. Are c and d scalars? I'm just not making any sense of what you are asking here.

-Dan

CaptainBlack

Well-known member
Ok give me a break, this is my first lesson in my new linear algebra book. Seems fairly straightforward but a little befuddled as to whether I am doing this. The question states "Describe geometrically (line, plane, or all of R^3) all linear combinations of...."
Then I have a matrix v = [1 2 3] w = [3 6 9]...... I do this.

cv-dw [-2 -4 -6] I believe it is R^3
v+w [4 8 11] Also R^3

If you could offer if I am still befuddled, [1 0 0] and [0 2 3]

forgot my proof....cv + dw = c[1 1 0]+d[0 1 1] it turns out as cv+dw = [c c+d d]

Need to start learning latex -.-

(Maybe I can get ahead of Jameson and his linear algebra *snickers*) Please use English when framing your question. Whatever you have used above has rendered your post incomprehensible.

If you do not understand what the question is asking post the question as asked.

CB

kanderson

Member
Sorry, I was really tired yesterday..

Describe geometrically (line, plane, or all of 3rd dimension (R^3)) all linear combinations of these two matrices... [1 2 3] and [3 6 9]

That was the question

Fantini

MHB Math Helper
Good night Kanderson. First you have to notice that $\begin{bmatrix} 3 & 6 & 9 \end{bmatrix}$ is a multiple of $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$. In fact, $\begin{bmatrix} 3 & 6 & 9 \end{bmatrix} = 3 \cdot \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$.

This means they generate the same space, therefore they are equivalent vectors.

Thus your space is generated by this sole vector, hence you have a dimension 1 space. A dimension 1 space is a line, and there's your answer. kanderson

Member
Oh I see..... Thank you.

Jameson

Staff member
Oh I see..... Thank you.
I'm not convinced that you do see, if you don't mind my saying Don't feel like you can't ask for more help. CB's comment was made so you can get help in the most efficient way. We want to help you.

Are you sure you understand the problem?

kanderson

Member
Well I see what he did there with the scalar...I understand that it fills a line...

but what about [1 0 0] and [0 2 3] I dont see any similarities....then
this one also [2 0 0] [0 2 2] [2 2 3]

I feel kind of narrow minded .......

Fantini

MHB Math Helper

In $\mathbb{R}^3$ we say that a 1-dimensional space is a line, a 2-dimensional space and a 3-dimensional space is $\mathbb{R}^3$ itself.

A linear combination of vectors $\alpha_1, \ldots, \alpha_n$ is the element $\gamma_1 \alpha_1 + \cdots + \gamma_n \alpha_n$.

For a space to be 1-dimensional there has to be only one generator, that is, every vector has to be a multiple of another (essentially, the linear combination of one vector).

For a space to be 2-dimensional there has to be two generators, that means every vector has to be a linear combination of those two.

In our cases, let us take the linear combination of those matrices. Denoting the scalars by $\alpha_1, \alpha_2$ we have that the linear combination is

$$\alpha_1 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} + \alpha_2 \begin{bmatrix} 3 & 6 & 9 \end{bmatrix}.$$

We have already noted that $\begin{bmatrix} 3 & 6 & 9 \end{bmatrix} = 3 \cdot \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$, therefore the combination becomes

$$\alpha_1 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} + \alpha_2 \begin{bmatrix} 3 & 6 & 9 \end{bmatrix} = \alpha_1 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} + 3 \alpha_2 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}.$$

Grouping the terms we have that

$$\alpha_1 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} + 3 \alpha_2 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = (\alpha_1 + 3 \alpha_2) \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = \gamma \begin{bmatrix} 1 & 2 & 3 \end{bmatrix},$$

where $\gamma$ is some scalar in $\mathbb{R}$. It doesn't matter that $\gamma = \alpha_1 + 3 \alpha_2$ but rather that it is a real scalar. Therefore, all elements that are linear combinations of those two matrices are, in fact, just multiples of the first. By our definition, this means it is a 1-dimensional space and hence a line.

I hope this clears up everything.